Given $u=x{{u}_{x}}+y{{u}_{y}}+\frac{1}{2}\left( u_{x}^{2}+u_{y}^{2} \right)$ , find a solution with $u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)$ .

Not confortrable with my solution as follows. Please help.

Standard Charpit's method leads to


$\Rightarrow p=a,q=b\Rightarrow u=ax+by+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$

$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right).$

$\Rightarrow {{b}^{2}}=\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax.$

$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+y\sqrt{\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax}\text{ where }a\text{ is an arbitary constant}.$

Solution alternative (for Clairaut's form) :


$z=ax+by+c\text{ }.$

$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+c\text{ }\Rightarrow \text{ }c=\frac{1}{2}\left( 1-{{x}^{2}} \right)-ax\text{ }.$

$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+by.$

  • $\begingroup$ No,it's alright.Actually ,this is how you derive Clairaut's form conditions. $\endgroup$ – Koro Nov 14 '15 at 3:41

Hint It is in Clairaut's form. Can you see?

$u=ax+by+\frac{1}{2}(a^2+b^2)$ then use the initial condition to eliminate $b$

  • $\begingroup$ A two-dimensional Clairaut's equation. So, is the Charpit's approach completely wrong? $\endgroup$ – Patrick Windance Nov 14 '15 at 3:35
  • $\begingroup$ No your approach is correct but it's more complicated, it is better to use Clairaut's for clear understanding. You can reach the result easily and clearly. $\endgroup$ – Kushal Bhuyan Nov 14 '15 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.